How do you solve $4tanx=2sec^2x$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-10-20T02:22:11

$x=pi/4$ and $x=(5pi)/4$

$(4tanx)/2=(2sec^2x)/2$

$2tanx=sec^2x$

Use the Pythagorean identity $sec^2x=tan^2x+1$

$2tanx=tan^2x+1$

$0=tan^2x-2tanx+1color(white)(aaa)$ Subtract $2tanx$ from both sides

$0=(tanx-1)(tanx-1)color(white)(aaa)$ Factor

$tanx-1=0color(white)(aaa)$ Set the factors equal to zero

$tanx=1color(white)(aaaaa)$ Add $1$ to both sides

For $0<=x<=2pi$ , $tanx=1$ where $sinx=cosx$
(because $tanx=sinx/cosx$ ).

Use the unit circle to find angles where
$sinx=cosx$ .

$tanx=1$ at $x=pi/4$ and $x=(5pi)/4$

How do you solve 4tanx=2sec^2x4tanx=2sec^2x for 0<=x<=2pi0<=x<=2pi?